#Do
apply(y,1,print)
#Note the space that is inserted before the "1." If you insert this space in
your function
apply(y,1,function(x){x<-unlist(x); if (!is.na(x[2]) & x[2]=='2k' &
!is.na(x[1]) & x[1]==' 1') 1 else 0} )
#you get the result you expect.
#Also, note that your !is.na conditions are redundant for the given example
because if the other conditions are true, the !is.na conditions default to
true, as well.
apply(y,1,function(x){x<-unlist(x); if (x[2]=='2k' & x[1]==' 1') 1 else 0} )
HTH,
Daniel
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